Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

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The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

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In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

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Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in JAVA):

Solution:
import java.util.*;
import java.io.*;
public class C {
    StringBuilder sb = new StringBuilder();
    void solve(BufferedReader in) throws Exception {
        int[] xx = toInts(in.readLine());
        int n = xx[0], m = xx[1];
        int[][] b = new int[n][m];
        for(int i = 0; i<n; i++) b[i] = toInts(in.readLine());
        int sz = 0;
        if(n>m) {
            sz += rm_col(b, n, m);
            sz += rm_row(b, n, m);
        } else {
            sz += rm_row(b, n, m);
            sz += rm_col(b, n, m);
        }
        if(notzero(b)) {
            System.out.println(-1);
        } else {
            System.out.println(sz);
            System.out.print(sb);
        }
    }
    int rm_col(int[][] b, int n, int m) {
        int ret = 0;
        for(int j = 0; j<m; j++) {
            int min = Integer.MAX_VALUE;
            for(int i = 0; i<n; i++) min = Math.min(min, b[i][j]);
            for(int i = 0; i<n; i++) b[i][j] -= min;
            for(int i = 0; i<min; i++) {
                sb.append("col " + (j+1)).append('\n');
            }
            ret += min;
        }
        return ret;
    }
    int rm_row(int[][] b, int n, int m) {
        int ret = 0;
        for(int i = 0; i<n; i++) {
            int min = Integer.MAX_VALUE;
            for(int j = 0; j<m; j++) min = Math.min(min, b[i][j]);
            for(int j = 0; j<m; j++) b[i][j] -= min;
            for(int j = 0; j<min; j++) {
                sb.append("row " + (i+1)).append('\n');
            }
            ret += min;
        }
        return ret;
    }
    boolean notzero(int[][] b) {
        for(int[] a: b) for(int v: a) if(v!=0) return true;
        return false;
    }
    int toInt(String s) {return Integer.parseInt(s);}
    int[] toInts(String s) {
        String[] a = s.split(" ");
        int[] o = new int[a.length];
        for(int i = 0; i<a.length; i++) o[i] = toInt(a[i]);
        return o;
    }
    void e(Object o) {
        System.err.println(o);
    }
    public static void main(String[] args) throws Exception{
        BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
        (new C()).solve(in);
    }
}